Spatial Sampling Requirements

How Fine Should the Grid Be?

We have determined what region of k-space we can observe and what resolution we can achieve. The remaining question is: what spatial grid should we use for discretizing the scene? Too coarse a grid aliases high-frequency content; too fine wastes computation. The Nyquist-Shannon sampling theorem, applied to k-space, gives the answer.

Theorem: Spatial Sampling Theorem for Imaging

If the k-space coverage is bounded within [κmax/2,κmax/2][-\kappa_{\max}/2, \kappa_{\max}/2] in each dimension (with κmax=22π(f0+W/2)/c\kappa_{\max} = 2 \cdot 2\pi(f_0 + W/2)/\text{c} for the maximum combined wavenumber), then the imaging grid spacing must satisfy

Δp2πκmax=λmin2,\boxed{\Delta p \leq \frac{2\pi}{\kappa_{\max}} = \frac{\lambda_{\min}}{2}},

where λmin=c/(f0+W/2)\lambda_{\min} = \text{c}/(f_0 + W/2) is the shortest wavelength in the system. Violating this condition causes spatial aliasing: replicas of the scene appear as ghost images.

This is the spatial analogue of the Nyquist theorem in time-frequency: you need at least two samples per wavelength in each direction to avoid aliasing. The maximum spatial frequency is κmax/2π\kappa_{\max}/2\pi, so the sampling rate must be at least κmax/π\kappa_{\max}/\pi samples per meter.

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Definition:

Angular Sampling Theorem

For a receiver array with aperture DD (total physical size), the angular sampling requirement is determined by the scene extent LL (diameter of the target region). The number of angular samples (independent Tx or Rx positions) needed is

Nangular2κLDπd(s,p0),N_{\text{angular}} \geq \frac{2\kappa L D}{\pi \cdot d(\mathbf{s}, \mathbf{p}_{0})},

where the factor accounts for the Fresnel zone structure. In the far field, this simplifies to: the angular spacing between adjacent Tx-Rx directions must be less than

Δθλ2L,\Delta\theta \leq \frac{\lambda}{2L},

to avoid grating lobes. This is the angular analogue of the spatial Nyquist condition.

Violating the angular sampling condition creates grating lobes — periodic replicas of the image in angle, analogous to the grating lobes of a phased array with element spacing >λ/2> \lambda/2.

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Example: Imaging Grid Design for a 10 GHz System

Design the imaging grid for a system with f0=10f_0 = 10 GHz, W=200W = 200 MHz, and a target region of size 16×1616 \times 16 m centered at range 20 m. Determine the grid spacing, number of voxels, and whether the standard 128×128128 \times 128 grid from the codebase is appropriate.

Grating Lobes and the Antenna Connection

As Caire points out in his note, the spatial sampling in imaging is intimately connected to the grating-lobe phenomenon in antenna arrays. An array with element spacing d>λ/2d > \lambda/2 produces grating lobes — spurious beams at angles satisfying the array factor periodicity. Similarly, an imaging grid with spacing Δp>λmin/2\Delta p > \lambda_{\min}/2 produces grating lobes in the reconstructed image — ghost replicas of the true scene.

The key insight from the 1D sinc/sampling analysis in Caire's paper: the total bandwidth determines the resolution (mainlobe width), while the sampling pattern determines the artifact structure (sidelobe/grating-lobe pattern). Irregular (non-periodic) sampling spreads grating-lobe energy over the image rather than concentrating it, which is why irregular arrays are generally preferred for imaging.

Spatial Aliasing in RF Imaging

Demonstrate how under-sampling the imaging grid creates aliasing artifacts. Compare the reconstructed image at different grid spacings to the Nyquist-sampled reference.

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🔧Engineering Note

Voxel Sizing Rules of Thumb

In practice, the voxel size is chosen as a compromise between computational cost and imaging quality:

  1. Each voxel should not span more than one range resolution cell: Δprc/(2W)\Delta p_r \leq \text{c}/(2W).
  2. Each voxel should not span more than one angular resolution cell of any receiver array.
  3. The total number of voxels QQ is limited by memory and computation: the sensing matrix ACMNK×Q\mathbf{A} \in \mathbb{C}^{MNK \times Q} requires 8MNKQ8 \cdot MNK \cdot Q bytes (complex64).
  4. Adaptive refinement (coarse grid first, refine around active voxels) can reduce QQ while maintaining resolution where it matters.

For the standard CommIT configuration (f0=10f_0 = 10 GHz, W=200W = 200 MHz, Q=1282Q = 128^2), each voxel is 12.5×12.512.5 \times 12.5 cm — much larger than the Nyquist limit but matched to the range resolution. This is deliberate: finer gridding would not improve imaging with 200 MHz bandwidth.

Common Mistake: Over-Sampling Does Not Improve Resolution

Mistake:

Using an extremely fine imaging grid (many more voxels than measurements) expecting that the finer grid will produce a higher-resolution image.

Correction:

Resolution is fundamentally limited by the k-space coverage (bandwidth and aperture). Over-sampling the grid beyond the Nyquist rate creates a smoother-looking image (sinc interpolation) but does not reveal features finer than Δr\Delta r or Δx\Delta x. Worse, a very fine grid makes the inverse problem more severely underdetermined (MNKQMNK \ll Q) and amplifies noise. The grid should be matched to the achievable resolution — typically ΔpΔr\Delta p \sim \Delta r in range and ΔpΔx\Delta p \sim \Delta x in cross-range.

Quick Check

A system has f0=28f_0 = 28 GHz and W=400W = 400 MHz. The Nyquist grid spacing for the imaging grid is approximately:

λmin/25.3\lambda_{\min}/2 \approx 5.3 mm

λ0/25.4\lambda_0/2 \approx 5.4 mm

c/(2W)=37.5c/(2W) = 37.5 cm

λ010.7\lambda_0 \approx 10.7 mm

Key Takeaway

The complete system design recipe for RF imaging is now clear: (1) Choose the array geometry to maximize angular diversity (k-space coverage). (2) Choose the bandwidth to achieve the desired range resolution: W=c/(2Δr)W = \text{c}/(2\Delta r). (3) Set the grid spacing to the achievable resolution (not the Nyquist limit, which is impractically fine). (4) Check the k-space coverage pattern for gaps. (5) Determine the SNR from the link budget. Every subsequent chapter in this book develops methods to solve the resulting inverse problem y=Ac+w\mathbf{y} = \mathbf{A}\mathbf{c} + \mathbf{w}.